验证数列$ \lim_{n \to \infty} \frac {1} {(n+n)^3} = 0 $的收敛性证明是否正确

Hey there! Let's walk through your proof together to check if it's solid.

Your core approach—using the formal definition of a sequence limit—is totally on the right track. The overall logic holds up, but we can tweak some details to make it more rigorous:

• First, starting with $|\frac{1}{(n+n)^3}| < \epsilon$: since $n$ is a positive integer (we're dealing with sequences here), the absolute value can be dropped immediately, leading to $\frac{1}{(2n)^3} < \epsilon$. This step is correct.
• Next, deriving the equivalent condition: $(2n)^3 > \frac{1}{\epsilon}$. Expanding this gives $8n^3 > \frac{1}{\epsilon}$, which simplifies to $n^3 > \frac{1}{8\epsilon}$, or $n > \sqrt[3]{\frac{1}{8\epsilon}} = \frac{1}{2\sqrt[3]{\epsilon}}$.
• The $N = \frac{1}{\epsilon}$ you chose is actually valid! Because when $\epsilon$ is small (which is the case we care about for limits), $\frac{1}{\epsilon}$ is larger than $\frac{1}{2\sqrt[3]{\epsilon}}$. So if $n > N$, it automatically satisfies $n > \frac{1}{2\sqrt[3]{\epsilon}}$, which makes $\frac{1}{(2n)^3} < \epsilon$ hold. The definition only requires some such $N$ to exist, not the smallest possible one, so your choice works.

That said, if you want to make the proof more precise, you could take $N = \max\left(1, \frac{1}{2\sqrt[3]{\epsilon}}\right)$—the $\max(1,...)$ ensures $N$ is a positive integer, which aligns with the fact that $n$ is a positive integer in the sequence. But even without that, your proof's logic is sound, and the conclusion that the limit equals 0 is correct.

备注：内容来源于stack exchange，提问作者Anqa012